Every not-orientable $n$-manifold embeds in an orientable $(n+1)$-manifold

Question

I have to prove that every not orientable smooth $n$-manifold $M$ can be embdedded in an orientable smooth $(n+1)$-manifold. My idea is to use the embedding theorem of Whitney but it just says that there exists an embedding $f: M\to\mathbb{R}^h$ for some $h\ge 1$. Can anyone help me, please?

Answer 1

Put $N:=\Lambda^nT^*M$ where $n=\dim(M)$, so there is a natural projection $p:N\to M$ and a point $y\in N$ with $p(y)=x$ is an $n$-linear alternating map $(T_xM)^n\to\mathbb R$. Then $T_yp:T_yN\to T_xM$ is surjective and the kernel of this map can be identified with $\Lambda^n(T_xM)^*$ itself. Now take a basis $v_1,\dots,v_n$ of $T_xM$ choose lifts $\tilde v_i\in T_yN$. Add the element in $\Lambda^n(T_xM)^*\cong ker(T_yp)$ that represents the unique multilinear map $\alpha$ such that $\alpha(v_1,\dots,v_n)=1$. Now consider the basis $\{\tilde v_1,\dots,\tilde v_n,\alpha\}$ for $T_yN$. A short computation (or a moment of thought) shows that starting from a different basis of $T_xM$, the resulting basis of $T_yN$ represents the same orientation. Thus you get a well defined orientation on $T_yN$ and it is easy to see that the resulting orientations are compatible, so this makes $N$ into an oriented manifold and then $M$ is embedded as the zero section.